# Limit of n*sin(1/n) as n goes to infinity

I have researched the question $\lim_{n \to \infty} n*\sin(\frac{1}{n})$ quite profusely, and I know that it equals to 1, and I know why:

A) You can use a change of variables and substitute, say, $m = \frac{1}{n}$ so that $m \to 0$ instead.

B) L'Hopital's rule

The problem is, we haven't used either of these methods in class, so I am wondering if there is any other possible way to approach this question?

• There's always Taylor expansion Commented Apr 6, 2016 at 22:42
• Haven't really learnt that either :P Commented Apr 6, 2016 at 22:43
• Have you used mean value theorem? Commented Apr 6, 2016 at 22:54
• @rtybase Pretty much all we've done is studied sequences and series (such as the ratio test, squeeze theorem etc.). The original question was actually to show whether the sum of $n*\sin(\frac{1}{n})$ diverges (or converges), and my approach was to show that because lim $n*sin(\frac{1}{n})$ doesn't equal 0, it must diverge. Commented Apr 6, 2016 at 22:59
• See some ideas here math.stackexchange.com/questions/75130/… Commented Apr 7, 2016 at 18:08

To prove $\sin x/x,\,\tan x/x\to 1$ as $x\to 0$, consider the areas of a small-angle sector of a circle and the right-angled triangles obtained by using a radius for a hypotenuse or base. The squeeze theorem will complete the proof since $\cos x \to 1$.